# Quantitative Aptitude: Inequalities Set 4 (New Pattern)

New Pattern Inequalities Questions IBPS PO, SBI PO Mains 2017, NIACL, NICL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

Directions: Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

1. 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
A) Quantity I > Quantity II
B) Quantity I â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option A
Solution
:
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
2. 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 â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option A
Solution
:
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
3. Quantity I: largest number if the average of 7 consecutive odd no. is 53
Quantity II: largest number if the avg of 6 consecutive odd no. is 56
A) Quantity I > Quantity II
B) Quantity I â‰¥Â  Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥Â  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option C
Solution
:
I: avg= x+ (n-1) [for consecutive odd/even number x= starting number, n= no of terms]
53= x+ 6 =>x=47 hence last digit = 47+ 2*(n-1)=47+12=59
II: 56=x+5 =>x=51 ; last number = 51+2*(n-1)=61
II>I
4. 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 â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥Â  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option C
Solution
:
I:Â  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
5. 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 â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥Â  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option A
Solution
:
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= 80Â  — 3
B= 240 = Â½Â Â  240
A= 240/2=120 days
6. Quantity I: If the price of an article increases by 20% then by how much % a family decrease its consumption to make expenditure same?
Quantity II: If the price of an article decrease by 14 (2/7) then by how much % a family increase its consumption to make expenditure same?
A) Quantity I > Quantity II
B) Quantity I â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option E
Solution
: I: 20%= 1/5 increase- for increase add the numerator to denominator
1/(1+5)=1/6= 16 (2/3)%
II: 14 (2/7)= 1/7 dec- for decrease subtract the numerator from denominator
1/(7-1)=1/6=16 (2/3)%
7. 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 â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥Â  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option C
Solution
:
I: A:B= 4:1 â€“ initially
A:B= 2:3 â€“ 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 â€“ 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
8. Quantity I: The length of the train if train crosses a pole in 30 sec and a platform of 300 m length in 60 seconds.
Quantity II: The length of a platform if 200 m long train crosses a pole in 25 sec and platform in 55 sec
A) Quantity I > Quantity II
B) Quantity I â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option A
Solution
:
I: Train : Platform = 30 : 30
300m =1 : 1= 300m
II: Train : Platform= 25:30=5:6
5=200 =>6=240
9. Quantity I: x, such that 3x2 â€“ 25x + 52 = 0
Quantity II: y, such that 2y2 â€“ 7y + 3 = 0
A) Quantity I > Quantity II
B) Quantity I â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option A
Solution
:
3x2 – 25x + 52 = 0
3x2 – 12x â€“ 13x + 52 = 0
Gives x = 4, 13/3
2y2 â€“ 7y + 3 = 0
2y2 â€“ 6y â€“ y + 3 = 0
So y = 1/2, 3
Put all values on number line and analyze the relationship
1/2â€¦ 3â€¦ 4â€¦ 13/3
10. Quantity I: x, such that 12x2 â€“ 5x â€“ 3 = 0
Quantity II: y, such that 3y2 â€“ 11y + 6 = 0
A) Quantity I > Quantity II
B) Quantity I â‰¥ Quantity II
C) Quantity II > Quantity I
D) Quantity II â‰¥ Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Â  Option E
Solution
:
12x2 â€“ 5x â€“ 3 = 0
12x2 + 4x â€“ 9x â€“ 3 = 0
Gives x = -1/3, 3/4
3y2 â€“ 11y + 6 = 0
3y2 â€“ 9y â€“ 2y + 6 = 0
Gives y = 2/3, 3
Put all values on number line and analyze the relationship
-1/3â€¦ 2/3â€¦ 3/4â€¦ 3

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