CAT 2018 Quant Questions with solutions (slot 2)


Question 1:
The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99 is
  1. 80707
  2. 80773
  3. 80730
  4. 80751
Option: 1

Nth term of the series can be written as
$\mathrm{tn}=(4 \mathrm{n}+3)(4 \mathrm{n}+7)$
$=16 \mathrm{n}^{2}+40 \mathrm{n}+21$
$\Sigma \mathrm{tn}=16 \Sigma \mathrm{n}^{2}+40 \Sigma \mathrm{n}+21 \Sigma 1$
$=16 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+40 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+21 \mathrm{n}$
here n = 23 (7, 11, 15….. 95 is an AP with common different 4 with 23 terms)
$\sum t_{n}=\frac{16 \times 23 \times 24 \times 47}{6}+20 \times 23 \times 24+21 \times 23$
= 80707

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Question 2:
How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
  1. 5
  2. 6
  3. 8
  4. 7
Option: 6

Let 'ab' be the two digit number. Where b \( \ne \) 0.

On interchanging the digits, the new number will be ‘ba’

As per the condition 10a+b > 3×(10b + a)

7a > 29b

For b = 1, a = {5, 6, 7, 8, 9}

For b = 2, a = {9}

For b = 3, no value of 'a' is possible.

Hence, there are a total of 6 such numbers


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Question 3:
The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is
Answer: 8
Given, $n^{3}-11 n^{2}+32 n-28>0$
When $n=2, n^{3}-11 n^{2}+32 n-28=0$
$\Rightarrow(n-2)\left(n^{2}-9 n+14\right)>28$
$\Rightarrow(n-2)(n-7)(n-2)>28$
For n < 2, (n – 2)(n – 7)(n – 2) is negative.
For 2 < n < 7, (n – 2)(n – 7)(n – 2) is negative.
For n > 7, (n – 2)(n – 7)(n – 2) is positive.
When n = 8, (n – 2)(n – 7)(n – 2) = 36, which is greater than 28. Least integral value of n which satisfies the inequation is 8.Online CAT 2021 Quant Course

Question 4:
Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.
Answer: 4000
Interest to be repaid to Ankit at the end of the year = 0.08X
Interest that Gopal would receive from Ishan in two cases are as given.
Case I: if he lends X + Y
Interest received = (X + Y) × 0.1 = 0.1X + 0.1Y
Interest retained by Gopal after paying to Ankit
= (0.1X + 0.1Y) – (0.08X) = 0.02X + 0.1Y
Given that Interest retained by Gopal is same as that accrued by Ankit
=> (0.02X + 0.1Y) = 0.08X
=> Y = 0.6X
Case II: if he lends X + 2Y
Interest received = (X + 2Y) × 0.1 = 0.1X + 0.2Y
Interest retained by Gopal after paying to Ankit
= (0.1X + 0.2Y) – (0.08X) = 0.02X + 0.2Y
Given that interest retained by Gopal would increase by 150
=> (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150
0.1Y = 150
=> Y = 1500 and X = 1500×0.6= 2500
Hence X + Y = 2500 + 1500 = 4000
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Question 5:
On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is
Answer: 50

Let 'x' and 'y' be the speed (in km/hr) of cars starting from both A and B respectively.
If they both move in east direction, then B will overtake A only if y > x.
Also, relative speed of both the cars when they move in east direction = (y - x) km/hr
It is mentioned that they take 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (y-x) km/hr.
Hence, (y - x) = 350/7 = 50 km/hr.Online CAT 2021 Quant Course

Question 6:
On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is
Answer: 24

Refer to the below diagram

Observe that triangle BPC and BQC are inscribed inside a semicircle. Hence,

\(\angle \) BPC = \(\angle \) BQC = 90°

Therefore, we can say that BQ \( \bot \) AC and CP \( \bot \) AB.

Also, In triangle ABC,

Area of triangle = (1/2)×Base×Height = (1/2) ×AB×CP = (1/2) ×AC×BQ

\( \Rightarrow \) BQ = \(\frac{{AB×CP}}{{AC}}\) = \(\frac{{30×20}}{{25}}\) = 24 cm.

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Question 7:
A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is
  1. 8
  2. 6√2
  3. 5√3
  4.  \(2\pi \)
Option: 3

Since $\Delta \mathrm{OAB}$ is equilateral,
radius of the circle is 5 $\mathrm{cm}$ .
In $\Delta \mathrm{OCD},$ by sine rule,
$\frac{5}{\sin 30^{\circ}}=\frac{\mathrm{CD}}{\sin 120^{\circ}}$
$\Rightarrow \mathrm{CD}=5 \frac{\sqrt{3}}{2} \times 2$
$=5 \sqrt{3}$

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Question 8:
Let f(x)=max{5x, 52-2x2}, where x is any positive real number. Then the minimum possible value of f(x) is
Answer: 20
Given x is positive real number. The minimum value of the maximum $\left\{5 x, 52-2 x^{2}\right\}$ will occur when both the graphs intersect. i.e., when $5 x=52-2 x^{2}$
$2 x^{2}+5 x-52=0$
$2 x^{2}+13 x-8 x-52=0$
$x(2 x+13)-4(2 x+13)=0$
$(x-4)(2 x+13)=0$
$x=4$ or $\frac{-13}{2}$
When $x=4, f(x)=20$Online CAT 2021 Quant Course

Question 9:
A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
  1. 52%
  2. 50%
  3. 55%
  4. 48%
Option: 2

Let the volume of the first and the second Solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture 

= (20 + 300S)

Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final Solution 

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = \(\frac{1}{2}\) = 50%

Hence, 50 is the correct answer.


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Question 10:
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ?lling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ?lled on Thursday if both pumps were used simultaneously all along?
  1. 4:36 pm
  2. 4:12 pm
  3. 4:24 pm
  4. 4:48 pm
Option: 3
Let x be the time, on a 24 hours clock, at which the tank is empty.
Time taken by pipe A alone to fill the tank is (20 – x) hrs.
Time taken by pipe B alone to fill the tank is (18 – x) hrs.
On the other day, A fill the tank for (15 – x) hrs and B for 2 hrs.
Let A and B be the rate of works of pipe A and B respectively.
$\Rightarrow(20-x) A=(18-x) B=(17-x) A+2 B$
$\Rightarrow \frac{A}{B}=\frac{2}{3}$
$\Rightarrow(20-x) 2=(18-x) 3$
$\Rightarrow(20-x) A=1$
Let $(20-x) A=1$
$A=\frac{1}{6}$
$B=\frac{1}{4}$
When both work simultaneously, time taken
$=\frac{1}{\frac{1}{6}+\frac{1}{4}}=2.4 \mathrm{hrs}=2 \mathrm{hrs} 24 \mathrm{min}$
The tank will be filled by $16 : 24 \mathrm{i.e.} .4 : 24 \mathrm{pm}$
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Question 11:
If a and b are integers such that 2x2 −ax + 2 > 0 and x2 −bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a−6b is
Answer: 36
$2 x^{2}-a x+2>0 \forall x \in R$
$\Rightarrow \Delta<0$
$\Rightarrow$ $a^{2}-4 \times 2 \times 2<0$
$\Rightarrow a^{2}<16$
$\Rightarrow-4$x^{2}-b x+8 \geq 0 \forall x \in R$
$\Rightarrow b^{2}-4(8) \leq 0$
$\Rightarrow-4 \sqrt{2} \leq b \leq 4 \sqrt{2}$
As b is integer $-5 \leq b \leq 5$
Therefore, maximum possible value of 2a – 6b is 2(3) – 6(–5) = 36Online CAT 2021 Quant Course

Question 12:
A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?
Answer: 48
Let the rate of filling of Type A and Type B pipes be a and b respectively.
Given 30×(10a + 45b) = 1 and 60×(8a + 18b) = 1
=> 30 × (10a + 45b) = 60 × (8a + 18b)
=> 10a + 45b = 16a + 36b
=> 3b = 2a or a = 1.5b
The total work = 30 × (10a + 45b) = 30 × (15b + 45b)
= 1800b
Required answer $=\frac{1800 b}{7 a+27 b}=\frac{1800 b}{10.5 b+27 b}=48$
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Question 13:
If N and x are positive integers such that NN = 2160 and N2 + 2N is an integral multiple of 2x, then the largest possible x is
Answer: 10
Given $\mathrm{N}^{\mathrm{N}}=2^{160}=2^{5 \times 32}$
$=\left(2^{5}\right)^{32} \Rightarrow \mathrm{N}^{\mathrm{N}}=32^{32}$
$\Rightarrow \mathrm{N}=32$
$\mathrm{N}^{2}+2^{\mathrm{N}}=32^{2}+2^{32}$
$\Rightarrow\left(2^{5}\right)^{2}+2^{32}$
$\Rightarrow 2^{10}+2^{32}$
$=2^{10}\left(1+2^{22}\right)$
Or $ \mathrm{x}$ is 10Online CAT 2021 Quant Course

Question 14:
Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals
Answer: 24
$t_{1}+t_{2}+\ldots+t_{n}=2 n^{2}+9 n+13 \rightarrow(1)$
$t_{1}+t_{2}+\ldots+\ {tn}-1=2(n-1)^{2}+9(n-1)+13 \rightarrow(2)$
From $(2)-(1),$ we get $t_{n}=\left(2 n^{2}+9 n+13\right)-\left(2(n-1)^{2}\right.$
$+9(n-1)+13 )=4 n+7$
Given $t_{k}=103=>4 k+7=103 \Rightarrow k=24 \quad$ Online CAT 2021 Quant Course

Question 15:
If p3 = q4 = r5 = s6, then the value of logs(pqr) is equal to
  1. 16/5
  2. 1
  3. 24/5
  4. 47/10
Option: 4
Let $p^{3}=q^{4}=r^{5}=s^{6}=k$
$p=k^{1 / 3}, q=k^{1 / 4}, r=k^{1 / 5}, s=k^{1 / 6}$
$\text{pqr}={{\text{k}}^{\left( \frac{20+15+12}{60} \right)}}={{\text{k}}^{\frac{47}{60}}}$
${{\log }_{\text{s}}}(\text{pqr})={{\log }_{{{\text{k}}^{\frac{1}{6}}}}}{{\text{k}}^{\frac{47}{60}}}$
$=\left(\frac{47}{60} \times 6\right) \log _{\mathrm{k}} \mathrm{k}$
$=\frac{47}{10}$
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Question 16:
Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his efficiency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?
  1. 13.5
  2. 11
  3. 12
  4. 14.5
Option: 1
Let r and g be the rates of work of Ramesh and Ganesh respectively.
Let $(\mathrm{r}+\mathrm{g}) 16=1$
$\Rightarrow(r+g)=\frac{1}{16}$
$(r+g) 7=\frac{7}{16}$
Remaining work to be done $=\frac{9}{16}$
Given, $(0.7 r+g) 10=\frac{9}{16}$
$7 r+10 g=9 r+9 g$
$g=2 r$
$r=\frac{1}{48}$
$g=\frac{1}{24}$
Time taken by g alone to complete the work $=\frac{\frac{9}{16}}{\frac{1}{24}}$=13.5
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Question 17:
A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
  1. 35.2
  2. 30.3
  3. 20.5
  4. 25.4
Option: 1

Final quantity of alcohol in the mixture = \(\frac{{700}}{{700 + 175}}×{(\frac{{90}}{{100}})^2}×[700 + 175]\) = 567 ml

Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml

Hence, the percentage of water in the mixture = \(\frac{{308}}{{875}} \times 100\) = 35.2 %


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Question 18:
In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is
Answer: 1098
Among a group of n persons, number of matches played = n(n – 1)/2
Among the Junior participants, let the number of girls be n.
The number of matches played among girls
= n(n – 1)/2 = 153
=> n(n – 1) = 306 = 18 × 17 => n = 18
Number of boys = 43 – 18 = 25
The number of matches played between a boy and a girl = 25×18 = 450
Among the Senior level participants, let the number of boys be n.
The number of matches played between two boys
= n(n – 1)/2 = 276
=> n(n – 1) = 552 = 24 × 23 => n = 24
The number of girls = 51 – 24 = 27
The number of matches played between a boy and a girl = 27 × 24 = 648
Required answer = 450 + 648 = 1098
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Question 19:
If A = {62n -35n -1: n = 1,2,3,...} and B = {35(n-1) : n = 1,2,3,...} then which of the following is true?
  1. Neither every member of A is in B nor every member of B is in A
  2. Every member of A is in B and at least one member of B is not in A
  3. Every member of B is in A.
  4. At least one member of A is not in B
Option: 2
$A=36^{n}-35 n-1=36^{n}-1^{n}-35 n$
Since $a^{n}-b^{n}$ is divisible by a $-b$ for all positive integral values of n, A is a multiple of 35 for any integral value of n and B is a set containing all the multiple of 35 including 0.
Hence, every member of A is in B but not every element of B is in A.

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Question 20:
A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?
  1. s≥6
  2. s≠6
  3. s≤6
  4. 5≤s≤7
Option: 1

>

We can see that area of parallelogram ABCD = 2×Area of triangle ACD

48 = 2×Area of triangle ACD 

Area of triangle ACD = 24

\((1/2)×CD×DA×sinADC = 24\)

\(AD×sinADC = 6\)

We know that \(sin\theta \) \( \le \) 1, Hence, we can say that AD \( \ge \) 6

\( \Rightarrow \) s \( \ge \) 6


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Question 21:
Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is
  1. 20
  2. 23
  3. 48
  4. 45
Option: 2

We want to maximize the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since a52 is 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3, ….a51.

The only way to do this is to assume that a2, a3…. a52 are in an AP with a common difference of 1.

Let the average of a2, a3…. a52 i.e. a27 be A.

(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

Since a52 = a27 + 25 and a52 = 100

=> A = 100 – 25 = 75

a2 + a3 + … + a52 = 75×51 = 3825

Given a1 + a2 +… + a52 = 52(A – 1) = 3848

Hence a1 = 3848 – 3825 = 23


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Question 22:
Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is
Answer: 5

Time taken to cover first 50 km at 100 km/hr = \(\frac{1}{2}\) hr.

Time taken to cover second 50 km at 50 km/hr = 1 hr.

Time taken to cover last 50 km at 25 km/hr = 2 hr.

When car 2 starts, car 1 has already covered 20 km.

So, time taken by car 1 to reach B after car 1 starts = total time - time required to travel first 20 km

= 3 hr 30 min - 12 min = 3 hr 18 min

Distance travelled by car 1 = (50 + 50 + 45) = 145 km

Distance from B = (150 - 145) km = 5 km

Hence, 5 is the correct answer.

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Question 23:
The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is
Answer: 105
Given $\frac{x+y+z}{3}=80$
$\Rightarrow x+y+z=240$ …. (1)
also $\frac{x+y+z+u+v}{5}=75$
$x+y+z+u+v=375$ …. (2)
From (1) and (2), $\mathrm{u}+\mathrm{v}=135$ …. (3)
$\frac{\mathrm{x}+\mathrm{y}}{2}+\frac{\mathrm{y}+\mathrm{z}}{2}=135$
$x+2 y+z=270$ …. (4)
From (1) & (4), y=30
$\Rightarrow x+z=210$
Since x ≥ z, x takes the minimum possible value at
x = 105Online CAT 2021 Quant Course

Question 24:
If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
  1. −32
  2. 48
  3. 64
  4. 16
Option: 3
Let a and b be the two numbers.
We know that for any two numbers $A M \geq G M$
$\Rightarrow \frac{a^{2}+b^{2}}{2} \geq a b$
$a b \leq \frac{97}{2}$
$a b \leq 48.5$
Among the options, only 64 is greater than 48.5

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Question 25:
For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is
  1. 9
  2. 7
  3. 6
  4. 8
Option: 7

P = {1,2,3,4} and Q = {2,3,5,6,}

PΔQ = {1, 4, 5, 6}

R = {1,3,7,8,9} and S = {2,4,9,10}

RΔS = {1, 2, 3, 4, 7, 8, 10}

(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}

Thus, there are 7 elements in (PΔQ)Δ(RΔS) .

hence, 7 is the correct answer.


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Question 26:
The smallest integer n for which 4n > 1719 holds, is closest to
  1. 33
  2. 37
  3. 39
  4. 35
Option: 3

\({4^n} > {17^{19}}\)

\( \Rightarrow \) \({16^{n/2}} > {17^{19}}\)

Therefore, we can say that n/2 > 19

n > 38


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Question 27:
The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is
  1. 2 : 5
  2. 1 : 3
  3. 1 : 4
  4. 3 : 10
Option: 2

Let 'a', 'b' and 'c' be the concentration of salt in Solutions A, B and C respectively.

It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting Solution has strength 20%.

\( \Rightarrow \) \(\frac{{a + 2b + 3c}}{{1 + 2 + 3}} = 20\)

\( \Rightarrow \) \(a + 2b + 3c = 120\) ... (1)

Also, if the proportion is 3 : 2 : 1, then the resulting Solution has strength 30%.

\( \Rightarrow \) \(\frac{{3a + 2b + c}}{{1 + 2 + 3}} = 30\)

\( \Rightarrow \) \(3a + 2b + c = 180\) ... (2)

From equation (1) and (2), we get

\( \Rightarrow \) \(b + 2c = 45\)

By observation, we find that b = c = 15 and a = 45.

So if we mix Solution B and C in any ratio we get the mixture with 15% concentration whereas A's strength = 45%.

Hence, the required ratio = \(\frac{{15}}{{45}}\) = 1 : 3


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Question 28:
The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio
  1. 1:4
  2. 2:9
  3. 1:3
  4. 3:8
Option: 1
Let the length and the breadth of the rectangle be L and B respectively.
Given that $\frac{\text { Area of rectangle }}{\text { Perimeter }^{2}}=\frac{1}{25} \Rightarrow \frac{L \times B}{(2(L+B))^{2}}=\frac{1}{25}$
$\Rightarrow 25 L B=4 L^{2}+4 B^{2}+8 L B$
$= L^{2}+B^{2}=(17 / 4) L B$
(Note: Alternatively, we can also solve the quadratic equation in terms of L/B and we’d get the same result, i.e. 4 or ¼ )
Since B < L, the ratio B : L = 1 : 4

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Question 29:
The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is
  1. 5 : 4
  2. 8 : 5
  3. 4 : 3
  4. 3 : 2
Option: 3
Given, ratio of the scores of Amal and Bimal is 11 : 14.
Let 11x and 14x be the scores of Amal and Bimal.
Let a be the score which is increased.
$\Rightarrow \frac{11 x+a}{14 x+a}=\frac{47}{56}$
$616 x+56 a=658 x+47 a$
$9 a=42 x$
$a=\frac{42 x}{9}$
Required ratio $=\left(14 x+\frac{42 x}{9}\right) : 14 x$
$=\left(1+\frac{1}{3}\right) : 1$
$=4 : 3$
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Question 30:
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
  1. 80 + 16π
  2. 86 + 8π
  3. 82 + 24π
  4. 88 + 12π
Option: 4

Area of the semicircle with AB as a diameter = \(\frac{1}{2}×\pi ×(\frac{{A{B^2}}}{4})\)

\( \Rightarrow \) \(\frac{1}{2}×\pi ×(\frac{{A{B^2}}}{4})\) = \(72×\pi \)

\( \Rightarrow \) \(AB = 24cm\)

It is also know that the area of the rectangle ABCD = 768 sq.cm

\( \Rightarrow \) AB×BC = 768

\( \Rightarrow \) BC = 32 cm

Observe that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

\( \Rightarrow \) 32+24+32+\(\pi ×24/2\)

\( \Rightarrow \) \(88 + 12\pi \)


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Question 31:
A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is
  1. 4 units
  2. 8 units
  3. 4√2 units
  4. 2√2 units
Option: 1

Since we want point A to be as close to the origin as possible, let point A lie on the x axis and its coordinates be (a, 0).
The distance of A from side BC (lying on the line x = 4) is the height of the triangle
=> The height of the triangle ABC = |a – 4|
Given the area of the triangle = 32
=> (1/2) × 8 × |a – 4| = 32 => |a – 4| = 8
=> a = 12 or –4
Required answer is the shortest distance from (0, 0) i.e. 4 when a = –4.
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Question 32:
There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio
  1. 229 : 141
  2. 220 : 149
  3. 239 : 161
  4. 251 : 163
Option: 3
Let the ratio of A and B in drum 2 be x : 1
Applying alligation,
$\frac{13}{20}-\frac{x}{x+1}=\frac{3}{4}=>\frac{13}{20}-\frac{x}{x+1}=\frac{3}{4} \times \frac{72-65}{100}=\frac{21}{400}$
$=>\frac{x}{25}-\frac{13}{20}=\frac{13}{20}-\frac{21}{400}=\frac{239}{400}$
$=>400 x=239 x+239$
$x=239 / 161$
Required ratio is 239 : 161.
Online CAT 2021 Quant Course

Question 33:
Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is
  1. 1 : 2
  2. 2 : 9
  3. 1 : 4
  4. 2 : 7
Option: 3

Car 3 meets car 1 at Q, which is 200 km from A.

Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.

As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.

Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.

As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1

And speed of car 2 : speed of car 3 = 1 : 2

So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2


Online CAT 2021 Quant Course

Question 34:
\(\frac{1}{{{{\log }_2}100}} - \frac{1}{{{{\log }_4}100}} + \frac{1}{{{{\log }_5}100}} - \frac{1}{{{{\log }_{10}}100}} + \frac{1}{{{{\log }_{20}}100}} - \frac{1}{{{{\log }_{25}}100}} + \frac{1}{{{{\log }_{50}}100}} = ?\)
  1. 1/2
  2. 0
  3. 10
  4. −4
Option: 1
We know that $\frac{1}{{{\log }_{b}}a}={{\log }_{a}}b$, therefore,
$\frac{1}{\log _{2} 100}-\frac{1}{\log _{4} 100}+\frac{1}{\log _{5} 100}-\frac{1}{\log _{10} 100}+\frac{1}{\log _{20} 100}-\frac{1}{\log _{25} 100}+\frac{1}{\log _{50} 100}$
= $\log _{100} 2-\log _{100} 4+\log _{100} 5-\log _{100} 10+\log _{100} 20-\log _{100} 25+\log _{100} 50$
=${{\log }_{100}}\left( \frac{2}{4}\times \frac{5}{10}\times \frac{20}{25}\times 50 \right)$
=${{\log }_{100}}10$
Using the relation ${{\log }_{{{a}^{m}}}}b=\frac{1}{m}{{\log }_{a}}b$
${{\log }_{100}}10={{\log }_{{{10}^{2}}}}10=\frac{1}{2}{{\log }_{10}}10=\frac{1}{2}$
Online CAT 2021 Quant Course

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