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2021 Questions QA
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2021 Complete Paper Solution | QA
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Question 1.
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The length of canvas 1.1 m wide required to build a conical tent of height 14m and the floor area 346.5 $m^{2}$ is
A
490 m
B
525 m
C
665 m
D
860 m
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Text Explanation:
Let's solve this step by step:
Area of floor = $346.5$ $m^2$
Floor is circular, so using area formula:
$346.5 = \pi r^2$
Radius of floor = $r = \sqrt{\frac{346.5}{\pi}} = 10.5$ m
Height of tent = $h = 14$ m
Slant height using Pythagoras theorem:
$l = \sqrt{h^2 + r^2} = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5$ m
Area of curved surface = $\pi r l$
$= \pi × 10.5 × 17.5 = 577.5$ m$^2$
Width of canvas = $1.1$ m
Length of canvas required = $\frac{577.5}{1.1} = 525$ m
The answer is $525$ m.
Question 2.
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In a cricket team of 11 players, the average age is 28 years. Out of these, the average ages of three groups of three players each are 25 years, 28 years, and 30 years respectively. If in these groups, the captain and the youngest player are not included and the captain is eleven years older than the youngest player, then the age of the captain is
A
33 years
B
34 years
C
35 years
D
36 years
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Text Explanation:
Given:
Players: 11
Team average: 28 years
Three groups of 3 players: $\frac{25+28+30}{3}$ years
Captain's age = Youngest player's age + 11
Let $x =$ youngest player's age
Average of 9 players $= \frac{3×25 + 3×28 + 3×30}{9} = \frac{249}{9}$
Using average formula:
$\frac{9 × \frac{249}{9} + x + (x+11)}{11} = 28$
$\frac{249 + 2x + 11}{11} = 28$
$249 + 2x + 11 = 308$
$2x = 48$
$x = 24$
Captain's age $= 24 + 11 = 35$ years
Question 3.
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At what angle the hands of a clock are inclined at 15 minutes past 5?
A
72.5 degrees
B
67.5 degrees
C
64 degrees
D
58.5 degrees
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Text Explanation:
Angle between the hour and minute hand $= |30H - \frac{11}{2}M|$
where $H$ is the hour hand, and $M$ is the minute hand
So,
$H=5$
$M = 15$
Substituting:
$= |30(5) - \frac{11}{2}(15)|$
$= |150 - \frac{165}{2}|$
$= |150 - 82.5|$
$= |67.5|$
Therefore, the angle between the hands is $67.5°$
Question 4.
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A man pays 40 times the annual rent to purchase a building. The rate % per annum he derives from his investment is
A
2.50%
B
3.33%
C
4.00%
D
6.67%
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Text Explanation:
Let the annual rent be $R$
Purchase price = $40R$
Rate of return = $\frac{\text{Annual Return}}{\text{Investment}} × 100\%$
$= \frac{R}{40R} × 100\%$
$= \frac{1}{40} × 100\%$
$= 2.50\%$
Therefore, the rate of return is 2.50% per annum.
Question 5.
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In a rhombus, the length of the two diagonals are 40m and 30m respectively. Its perimeter will be
A
100 m
B
70 m
C
80 m
D
90 m
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Text Explanation:
Given:
Diagonal 1 $(d_1)$ = 40m
Diagonal 2 $(d_2)$ = 30m
In a rhombus, using diagonals to find side $(s)$:
$s^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2$
$s^2 = (\frac{40}{2})^2 + (\frac{30}{2})^2$
$s^2 = 400 + 225$
$s^2 = 625$
$s = 25$ meters
Perimeter of rhombus = $4s$
$= 4 × 25$
$= 100$ meters
Therefore, the perimeter is $100$ meters.
Question 6.
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The HCF of $(\frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \frac{21}{40})$ is
A
$\frac{3}{5}$
B
$\frac{252}{5}$
C
$\frac{63}{700}$
D
$\frac{3}{1400}$
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Text Explanation:
For fractions $\dfrac{a_1}{b_1}, \dfrac{a_2}{b_2}, ..., \dfrac{a_n}{b_n}$, the HCF formula is $\dfrac{HCF(a_1, a_2, ..., a_n)}{LCM(b_1, b_2, ..., b_n)}$
Given fractions are $\frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \frac{21}{40}$.
HCF of numerators (9, 12, 18, 21) = 3, and LCM of denominators (10, 25, 35, 40) = 1400.
Therefore, the HCF: $\frac{3}{1400}$
Question 7.
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The odd term out in the series 15, 16, 34, 105, 424, 2124, 12756 is
A
105
B
16
C
2124
D
34
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Text Explanation:
The series goes like this $=(\text{the number}\times n)+n$
to give the next number, where $n$ is consecutive first natural numbers.
$(15\times1)+1=16$
$(16\times2)+2=34$
$(34\times3)+3=105$
$(105\times4)+4=424$
$(424\times5)+5=2125$
Hence, $2124$ is the odd term out in the above series.
Question 8.
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$(\frac{4}{7})$ of a pole is in the mud. When $(\frac{1}{3})$ of it is pulled out, 250 cm of the pole is still in the mud. The length of the pole is
A
1050 cm
B
1350 cm
C
437.5 cm
D
750 cm
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Text Explanation:
Let total length of pole be $x$ cm
Initially, $\frac{4}{7}x$ is in mud.
After pulling out $\frac{1}{3}$ of the pole, $250$ cm remains in mud.
Therefore:
$\frac{4}{7}x - \frac{1}{3}x = 250$
$\frac{12}{21}x - \frac{7}{21}x = 250$
$\frac{5}{21}x = 250$
$x = 250 × \frac{21}{5}$
$x = 1050$ cm
Therefore, the total length of the pole is 1050 cm.
Question 9.
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A is working, and B a sleeping partner in a business. A puts in Rs.12000 and B Rs. 20000. A receives 10% of the profit for managing, the rest being divided in proportion to their capitals. Out of a total profit of Rs. 9600, the money received by A will be
A
Rs. 3240
B
Rs. 3600
C
Rs. 4200
D
Rs. 4840
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Text Explanation:
Given:
Total profit = ₹$9600$
A's capital = ₹$12000$
B's capital = ₹$20000$
A gets $10\%$ of profit for managing
Rest divided in proportion to capital
A's managing share = $10\%$ of $9600$ = ₹$960$
Remaining profit = $9600 - 960$ = ₹$8640$
Proportion for remaining profit:
A : B = $12000 : 20000 = 3 : 5$
A's share from remaining = $\frac{3}{8} × 8640$ = ₹$3240$
Total money A receives = Managing share + Proportional share
= $960 + 3240$ $=$ ₹$4200$
Therefore, A will receive ₹$4200$.
Question 10.
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A pump can be operated both for filling a tank and for emptying it. The capacity of the tank is 2400 $m^{3}$. The emptying capacity of the pump is 10 $m^{3}$ per minute higher than its filling capacity. Consequently, the pump needs 8 minutes less to empty the tank as to fill it. The filling capacity of the pump is
A
40 $m^3$/minute
B
50 $m^3$/minute
C
60 $m^3$/minute
D
80 $m^3$/minute
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Text Explanation:
Let filling capacity = $x$ $m^3$/minute
Then emptying capacity = $(x + 10)$ $m^3$/minute
Given tank volume = $2400$ $m^3$
For filling: $\frac{2400}{x} = t$ minutes
For emptying: $\frac{2400}{x+10} = (t-8)$ minutes
Therefore:
$\frac{2400}{x} - \frac{2400}{x+10} = 8$
$\frac{2400(x+10) - 2400x}{x(x+10)} = 8$
$\frac{24000}{x(x+10)} = 8$
$24000 = 8x(x+10)$
$24000 = 8x^2 + 80x$
$8x^2 + 80x - 24000 = 0$
$x^2 + 10x - 3000 = 0$
Solving quadratic:
$x = 50$ or $x = -60$ (reject negative)
Therefore, filling capacity is 50 $m^3$/minute.
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