CUET UG Mathematics Applied Logical Ability And Logical Reasoning Previous Year Questions (PYQs)
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Match List I with List II
Choose the correct answer from the options given below:
| List I | List II |
| $3^3 \equiv b \pmod{9}$ | 4 |
| $2^5 \equiv b \pmod{15}$ | 0 |
| $4^3 \equiv b \pmod{10}$ | 2 |
| $5^3 \equiv b \pmod{12}$ | 5 |
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2
3
4
Solution
A. $3^3=27$
$27 \equiv 0 \pmod{9}$
So A → II
B. $2^5=32$
$32 \equiv 2 \pmod{15}$
So B → III
C. $4^3=64$
$64 \equiv 4 \pmod{10}$
So C → I
D. $5^3=125$
$125 \equiv 5 \pmod{12}$
So D → IV
Correct matching:
A-II, B-III, C-I, D-IV
A mixture contains milk and water in the ratio $8:x$.
If 3 litres of water is added in 33 litres of mixture, the ratio of milk and water becomes $2:1$, then value of $x$ is:
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Solution
Initial ratio = $8:x$
Total mixture = 33 litres
Milk = $\frac{8}{8+x}\times 33$
Water = $\frac{x}{8+x}\times 33$
After adding 3 litres water:
Milk remains same
Water becomes
$\frac{x}{8+x}\times 33 + 3$
Given final ratio:
$\frac{\text{Milk}}{\text{Water}}=\frac{2}{1}$
So,
$\frac{\frac{8}{8+x}\times 33}{\frac{x}{8+x}\times 33 + 3}=2$
Multiply by $(8+x)$:
$\frac{264}{33x + 3(8+x)}=2$
$264=2(33x + 24 + 3x)$
$264=2(36x+24)$
$264=72x+48$
$216=72x$
$x=3$
A motorboat can travel in still water at the speed $15 \text{ km/h}$, while the speed of the current is $3 \text{ km/h}$.
Time taken by boat to go $36 \text{ km}$ upstream is:
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Solution
Upstream speed
$= 15-3=12 \text{ km/h}$
Time
$=\frac{\text{Distance}}{\text{Speed}}$
$=\frac{36}{12}$
$=3 \text{ hr}$
Hari covers $100 \text{ m}$ distance in $36$ seconds. Ram covers the same distance in $45$ seconds.
In a $100 \text{ m}$ race, Hari ahead from Ram is:
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Solution
Speed of Hari
$=\frac{100}{36}=\frac{25}{9} \text{ m/s}$
Speed of Ram
$=\frac{100}{45}=\frac{20}{9} \text{ m/s}$
In 36 seconds (when Hari finishes race):
Distance covered by Ram
$=\frac{20}{9}\times 36$
$=80 \text{ m}$
So Hari wins by
$100-80=20 \text{ m}$
A pipe can empty $\left(\frac{5}{6}\right)$th part of a cistern in 20 minutes.
The part of cistern which will be empty in 9 minutes is:
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Solution
Pipe empties $\frac{5}{6}$ part in 20 minutes.
Rate per minute
$=\frac{5}{6} \div 20$
$=\frac{5}{6}\times \frac{1}{20}$
$=\frac{5}{120}$
$=\frac{1}{24}$
So in 1 minute it empties $\frac{1}{24}$ part.
In 9 minutes:
$=\frac{9}{24}$
$=\frac{3}{8}$
Sitaram, a money lender lent a part of Rs. $200000$ to Shyam at simple interest $6%$ p.a. and the remaining to Sushil at $10%$ p.a. at simple interest.
Sitaram earned an annual interest income of Rs. $18000$.
In what ratio did Sitaram lent the money at $6%$ p.a. and $10%$ p.a. respectively?
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Solution
Let amount lent at $6%$ be $x$.
Then amount lent at $10%$
$=200000-x$
Total interest:
$0.06x+0.10(200000-x)=18000$
$0.06x+20000-0.10x=18000$
$20000-0.04x=18000$
$0.04x=2000$
$x=50000$
So,
Amount at $6%$ = $50000$
Amount at $10%$ = $150000$
Ratio:
$50000:150000$
$=1:3$
Sitaram, a money lender lent a part of Rs. $200000$ to Shyam at simple interest $6%$ p.a. and the remaining to Sushil at $10%$ p.a. at simple interest.
Sitaram earned an annual interest income of Rs. $18000$.
How much money did Shyam borrow?
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Solution
From previous calculation,
Amount lent at $6%$ = Rs. $50000$
Shyam borrowed at $6%$.
Therefore,
Shyam borrowed Rs. $50000$
Sitaram, a money lender lent a part of Rs. $200000$ to Shyam at simple interest $6%$ p.a. and the remaining to Sushil at $10%$ p.a. at simple interest.
Sitaram earned an annual interest income of Rs. $18000$.
What amount of money is lent at $10%$ p.a. simple interest?
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Solution
From earlier calculation:
Amount at $6%$ = Rs. $50000$
Total principal = Rs. $200000$
So amount at $10%$
$=200000-50000$
$=150000$
Sitaram, a money lender lent a part of Rs. $200000$ to Shyam at simple interest $6%$ p.a. and the remaining to Sushil at $10%$ p.a. at simple interest.
Sitaram earned an annual interest income of Rs. $18000$.
What is the ratio of the interest paid by Shyam and Sushil respectively?
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Solution
$50000\times 6% = 3000$
Interest from Sushil:
$150000\times 10% = 15000$
Ratio:
$3000:15000$
$=1:5$
CUET UG Mathematics Applied
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and More.
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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