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CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Logical Ability And Logical Reasoning PYQ

CUET UG Mathematics Applied PYQ
Match List I with List II
 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
Choose the correct answer from the options given below:





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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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 $(8+x)$:

$\frac{264}{33x + 3(8+x)}=2$

$264=2(33x + 24 + 3x)$

$264=2(36x+24)$

$264=72x+48$

$216=72x$

$x=3$
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Upstream speed
$= 15-3=12 \text{ km/h}$

Time
$=\frac{\text{Distance}}{\text{Speed}}$

$=\frac{36}{12}$

$=3 \text{ hr}$
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$100-80=20 \text{ m}$
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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:

$=\frac{9}{24}$

$=\frac{3}{8}$
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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:

$50000:150000$

$=1:3$
CUET UG Mathematics Applied PYQ
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From previous calculation,

Amount lent at $6%$ = Rs. $50000$

Shyam borrowed at $6%$.

Therefore,

Shyam borrowed Rs. $50000$
CUET UG Mathematics Applied PYQ
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?





Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From earlier calculation:

Amount at $6%$ = Rs. $50000$

Total principal = Rs. $200000$

So amount at $10%$

$=200000-50000$

$=150000$
CUET UG Mathematics Applied PYQ
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?





Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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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