# Hard Level Inequality in Quantitative Aptitude Section for SBI Mains Exams | Day 15

__Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them__**.**

**Q1. Quantity I: Sum, if the difference between Compound Interest and Simple Interest at 20% rate of interest in 2 years is Rs 800**

Quantity II: Sum, if the compound interest of 2 years at the rate in which a sum of Rs 125 becomes Rs 216, in 3 years is Rs 6600

Quantity II: Sum, if the compound interest of 2 years at the rate in which a sum of Rs 125 becomes Rs 216, in 3 years is Rs 6600

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q2. Quantity I: Cost price, if after selling article of Rs 25,000 Marked price at 20% discount a man gains 33 (1/3)%.**

**Quantity II: Cost Price, if after selling article of Rs 16,000 marked price at 25% discount a man gains 20%**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q3. Quantity I: distance if a person goes to a distance at 80 kmph he reaches 15 min earlier and if he goes their at 70 kmph he reaches 15 mins late.**

**Quantity II: distance if two buses travels in the direction of each other at speed of 60 kmph and 50 kmph. When they meet, it was found that faster bus travelled 30 km more than the slower bus.**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q4. Quantity I: Number of days in which A can alone do the work given A and B can do a work in 80 days, B and C can do a work in 60 days, C and A can do work in 96 days.**

**Quantity II: Number of days in which A can alone do the work given A and B together can complete a work in 80 days. They work together for 10 days then A left the work. If B did the remaining work in 210 days.**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q5. Quantity I: A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10 Litre of the mixture is taken out and 10 Litre of liquid B was poured into the jar the ratio became 2:3. The quantity of liquid A contained in the jar initially is?**

**Quantity II: A jar contained a mixture of two liquid A and B in the ratio 6:5. When 10 Litre of mixture is taken out and 10 Litre of liquid B was poured into the jar , the ratio became 3:5. The quantity of liquid B contained in the jar finally.**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q6.The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.**

Quantity I: Cost Price of an article if it is sold making a loss of 10% given that if the cost price was 20% less, a profit of Rs 12 could be made.

Quantity II: Labeled Price of an article if a discount of 30% is given on it and then sold for Rs 105 at a loss of 20%

Quantity I: Cost Price of an article if it is sold making a loss of 10% given that if the cost price was 20% less, a profit of Rs 12 could be made.

Quantity II: Labeled Price of an article if a discount of 30% is given on it and then sold for Rs 105 at a loss of 20%

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q7**

**. Quantity I: Speed of boat in still water f it took 24 hours to cover a distance of 64 km going downstream and back. Given the speed of current to be 2 km/hr**

Quantity II:

Quantity II:

**Average speed of a boy to go from home to school if his average speed is 2 km/hr from school to home and he covers a total of 12 km in 5 hours.**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q8. Quantity I: A and B together can do a piece of work in 4 days. If A alone can do the same work in 6 days, then B alone can do the same work in?**

**Quantity II: A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q9. Quantity I: The age of P is twelve times that of her daughter Q. If the age of Q is 3 years, what is the age of P?**

**Quantity II: The ratio between the present ages of A and B is 2:3. 4 years ago the ratio between their ages was 5:8. What will be A’s age after 7 years?**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Q10. There are 5 Brown balls, 4 Blue balls & 3 black balls in a bag .Four balls are chosen at random**

Quantity I: The probability of their being 2 Brown and 2 Blue ball Quantity

Quantity I: The probability of their being 2 Brown and 2 Blue ball Quantity

**Quantity II: The probability of their being 2 Brown, 1 Blue & 1 blacks**

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

E) Quantity I = Quantity II or Relation cannot be established

**Solution:**

**Q1, SOL: A**

Quantity I: SI for 2 years =2*20%=40%

CI for 2 years =20+20+20*20/100=44%

diff=44-40=4%

4%=800 =>100%=2000

Quantity II: First find rate

Cube root (125): Cube root(216)=5:6 (cube root as year=3)

(6-5)/5*100=20%

CI for 2 years =44%

44%=6600

100%=15000

I>II

CI for 2 years =20+20+20*20/100=44%

diff=44-40=4%

4%=800 =>100%=2000

Quantity II: First find rate

Cube root (125): Cube root(216)=5:6 (cube root as year=3)

(6-5)/5*100=20%

CI for 2 years =44%

44%=6600

100%=15000

I>II

**Q2. SOL: A**

I: gain=33(1/3)% = 1/3

CP:SP=3:4

discount=20%=1/5

MP:SP=5:4

cp:sp:mp= 3:4:5

5=25000 hence 3= 3/5*25000=15000

II: cp:sp:mp=5:6:8

8=16000 hence 5=1000

CP:SP=3:4

discount=20%=1/5

MP:SP=5:4

cp:sp:mp= 3:4:5

5=25000 hence 3= 3/5*25000=15000

II: cp:sp:mp=5:6:8

8=16000 hence 5=1000

**Q3. SOL: C**

I:
D=(S1*S2)/(S1- S2) * (T1+T2)/60

solving we get D=280 Km

II. Difference in speed = 10 kmph; means in 1 hour the difference of distance covered is 10 km; so 30 km difference will be in 3 hours. So total distance=60*3 + 50*3=330 km

II>I

solving we get D=280 Km

II. Difference in speed = 10 kmph; means in 1 hour the difference of distance covered is 10 km; so 30 km difference will be in 3 hours. So total distance=60*3 + 50*3=330 km

II>I

**Q4. SOL: A**

I: A+B=80 —— 6 (total work=480)

B+C=60 —— 8

C+A=96 ——5

2 (A+B+C)= 19

A+B+C=19/2

A= 19/2-8=3/2

480*2/3=320 days

II: Let total work = 80

1 day work of A and B= 1

10 ——- 1

remaining= 80-10=70

B did 70 work in 210 days

hence 80 work in 240

A+B= 80 — 3

B= 240 = ½ 240

A= 240/2=120 days

B+C=60 —— 8

C+A=96 ——5

2 (A+B+C)= 19

A+B+C=19/2

A= 19/2-8=3/2

480*2/3=320 days

II: Let total work = 80

1 day work of A and B= 1

10 ——- 1

remaining= 80-10=70

B did 70 work in 210 days

hence 80 work in 240

A+B= 80 — 3

B= 240 = ½ 240

A= 240/2=120 days

**Q5. SOL: C**

I: A:B= 4:1
– initially

A:B= 2:3 – finally

Make Both A equal in the two equation as only B has been added so multiply eq 2 by 2, we get

A:B= 4:6;

difference of B=6-1=5 =10 L

=>1=2

hence total=(4+6)*2=20 L

water in intial =4/5*20=16 L

II: A:B= 6:5 – intital

A:B= 3:5 final

to make A same, multiply eq 2 by 2

A:B= 6:10

difference in B=10-5=5 this is equal to 10L =>1=2

6=12

B=10=20

Hence

16<20

I<II

A:B= 2:3 – finally

Make Both A equal in the two equation as only B has been added so multiply eq 2 by 2, we get

A:B= 4:6;

difference of B=6-1=5 =10 L

=>1=2

hence total=(4+6)*2=20 L

water in intial =4/5*20=16 L

II: A:B= 6:5 – intital

A:B= 3:5 final

to make A same, multiply eq 2 by 2

A:B= 6:10

difference in B=10-5=5 this is equal to 10L =>1=2

6=12

B=10=20

Hence

16<20

I<II

**Q6. SOL: E**

I: Let CP = Rs 100, then at 10%
loss, SP = Rs 90

Now if CP is 20% less, means CP = Rs 80, then profit is = 90 – 80 = Rs10

but given profit is Rs 12

So if profit is Rs 10, then CP = Rs100

If profit is Rs 12, then CP is 100/10 * 12 = Rs 120

II: Use formula MP = (100-loss%)/(100-dicount%) * CPS

So MP = (100-20)/(100-30) * 105 = Rs 120

Hence I = II

Now if CP is 20% less, means CP = Rs 80, then profit is = 90 – 80 = Rs10

but given profit is Rs 12

So if profit is Rs 10, then CP = Rs100

If profit is Rs 12, then CP is 100/10 * 12 = Rs 120

II: Use formula MP = (100-loss%)/(100-dicount%) * CPS

So MP = (100-20)/(100-30) * 105 = Rs 120

Hence I = II

**Q7. SOL: A**

I: Let speed of boat = x
km/hr

So 64/(x+2) + 64/(x-2) = 24

Solve, x = 6 km/hr

II : Distance from home to school is 12/2 = 6 km

Let speed from home to school is x km/hr

So using distance = total time * average speed

6 = 5 *[2*x/(2+x)]

Solve, x = 3 km/hr

Hence I > II

So 64/(x+2) + 64/(x-2) = 24

Solve, x = 6 km/hr

II : Distance from home to school is 12/2 = 6 km

Let speed from home to school is x km/hr

So using distance = total time * average speed

6 = 5 *[2*x/(2+x)]

Solve, x = 3 km/hr

Hence I > II

**Q8. SOL: E**

I. B work= 1/4 – 1/6 = 2/24 ==> 12 days

II. A’s 1hr work 1/4 .

(B+C’s) 1 hr work 1/3.

(A+C’s) 1hr work 1/2.

A+B+C 1hr work = 1/4+1/3=7/12.

B’s work =7/12 – 1/2=1/12

12hours.

(B+C’s) 1 hr work 1/3.

(A+C’s) 1hr work 1/2.

A+B+C 1hr work = 1/4+1/3=7/12.

B’s work =7/12 – 1/2=1/12

12hours.

**Q9. SOL: A**

**I**

Ratio P:Q 12:1

1…….== 3

12 ? == 12*3 =36years.

**II**

(5x+4)/(8x+4) =2/3

15x+12 = 16x+8

x=4.

A’s age 4 yrs ago 5*4=20

Then A’s age after 7yrs is 20+4+7=31yrs.

**Q10. SOL: C**

I. (5c2 * 4c2)/12c4 = 60/495=4/33

II. (5c2 * 4c1 *3c1)/12c4 =120/95=8/33

II. (5c2 * 4c1 *3c1)/12c4 =120/95=8/33

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