# Crack IBPS PO : Data Sufficiency – Aptitude Day 70

D.1-5) In the following question two quantities are given Quantity 1 and quantity 2 By solving those quantities give corresponding answer.

Q.1) Quantity 1: The average height of 16 students in a class is 142 cm. If the height of the teacher is added the average increases by 1 cm. What is the height of the teacher ?

Quantity 2: A spiral is made up of 13 successive semicircles, with centres alternately at A and B, starting with centre at A. The radii of semicircles thus developed are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm and so on. What is the total length of the spiral

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Quantity 1:

Given, average height of 16 students = 142 cm

Total height = 142 x 16 cm

and, if the height of the teacher is added the average, then

The average height of 16 students + Teacher = 143

Total height = 17 x 143 cm

Hence, Height of the teacher = (17 x 143) – (142 x 16) = 159 cm

Quantity 2:

Circumference of first semicircle =πr=0.5π

Circumference of first semicircle =πr=1π

Circumference of first semicircle =πr=1.5π

Number of semicircles=13

The length of the spiral=n/2 (2a+(n-1)d)

a=0.5π,d=0.5π

=13/2 (2×0.5π+(13-1)0.5π)

=13/2×7π

=13/2×7×22/7

=13×11=143cm

Hence Quantity I >Quantity II

Q.2) Quantity 1: In how many different ways the letters of the word BLOATING be arranged?

Quantity 2: In how many different ways the letters of the word SOLITARY be arranged?

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

b)

Quantity 1:

There are 8 letters in the given word BLOATING and none of the letters comes more than once.

The required number of ways = 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

Quantity 2:

There are 8 letters in the given word SOLITARY and none of the letters comes more than once.

The required number of ways = 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

Q.3) Quantity 1: Average score of Rahul, Manish & Suresh is 63. Rahul’s score is 15 less than Ajay and 10 more than Manish. If Ajay scored 30 marks more than the average score of Rahul, Manish & Suresh, what is the sum of Manish’s and Suresh’s scores?

Quantity 2: 400 persons working 9 hours per day complete 1/4th of the work in 10 days. The number of additional persons working 8 hours per day, required to complete the remaining work in 20 days?

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

e)

Quantity 1:

Let the score of Ajay = x

then, Rahul’s score = x – 15

and, Manish’s score = x – 25

According to the question,

Ajay’s score = 63 + 30 = 93

Hence, Rahul’s score = 93 – 15 = 78

Manish’s score = 93 – 25 = 68

Total marks of Rahul, Manish & Suresh = 63 x 3 = 189

Then, Suresh’s score = 189 – (78 + 68) = 43

The sum of Manish’s and Suresh’s scores = 68 + 43 = 111

Quantity 2:

400 person complete 1/4th of the work in 9×10 hours = 90 hours

= 400 person complete the work in 90 × 4 = 360 hours

=1 person completes the work in 360 × 400 hours

Let the number of persons completing the work in 20 days be x.

Remaining work to be done = 1 – 1/4 = 3/4

x person complete 3/(4 ) th of the work in 8×20 hours = 160 hours

= x person complete the work in 160×4/3 hours

= 1 person completes the work in 160×4/3×x hours

Time taken by 1 person to complete the work in both the cases will be same

So, 360 × 400 = 160×4/3×x

x = 675

So additional people required to complete the remaining work = 675 – 400 = 275

Hence Quantity I < Quantity II

Q.4) Quantity 1: A shopkeeper bought a bat and sold it at a loss of 15 %. If he had bought it for 20 % less and sold it for Rs. 147.2 more, he would have had profit of 35 %. What is the cost price of bat ?

Quantity 2: A sells a bicycle to B at a profit of 20% and B sells it to C at a profit of 25%.If C pays Rs.225 for it, what did A pay for it.

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Quantity 1:

Let cost price of bat is x.

Then S.P. =85x/100=17x/20.

If he had bought it for 20 % less:

C.P. =80x/100=4x/5

And, S.P. =17x/20+ 147.2 =(17x + 2944)/20

Now,

(S.P. – C.P.)/C.P × 100 = 35

[(17x + 2944)/20-4x/5]/((4x/5) )× 100 = 35

= (17x + 2944 – 16x)/(20 )×5/4x × 100 = 35

=(x + 2944)/20×125/x= 35

=( x + 2944)/4×25/x= 35

= 25x + 73600 = 140x

= x =73600/115= 640

Cost price of Bat = Rs 640

Quantity 2:

C.P of A =225×100/(100+20)×100/(100+25)

=225×100/120×100/125=150

Rs.150

Hence Quantity 1>Quantity2

Q.5) Quantity 1: After working for 8 days .Anil finds that only 1/3 of the work has been done.He employs Rakesh who is 60% efficient as Anil.How many more days will Anil take to complete the job

Quantity 2: 8 men and 4 women together can complete a piece of work in 6 days. Work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days, 4 men left and 4 new women joined, in how many more days will the work be completed ?

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Explanation:

Quantity 1:

In 8 days Anil does=1/3 rd work

In 1 day=1/24 thwork

Rakesh’s one day work=60% of 1/24=1/40 thwork

Remaining work=1-1/3=2/3

Anil & Rakesh’s one day’s work

=1/24+1/40=1/15 th

2/3 rd work=15×2/3=10 days

Quantity 2:

Work completed by 8M + 4W in 2 days = 2 ×1/6=1/3

Work done by 1 man = work done by 2 women

8 men + 4 women = (16 women + 4 women) = 20 women

8 men + 4 women = (8 men + 2 men) = 10 men

Work done by 20 women in 1 day = 1/20× 6 =1/120

Work done by 10 men in 1 day =1/10× 6 =1/60

As 4 men left and 4 new women joined, so 4M + 8W = 4 ×1/60+ 8 ×1/120=1/15+1/15=2/15

Work to be done = 1 -1/3=2/3

No of days required =(2/3)/(2/15)= 5 days.

D.6-10) In the following question two quantities are given Quantity 1 and quantity 2 By solving those quantities give corresponding answer.

Q.6) Quantity 1:

21.16% of 2000+46% of 350×(92÷11.5)=?

Quantity 2:

22234+32×1.5=?+18999-1747×2-1588+6622

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

e)

Quantity 1:

423.2+ (161×8)=1711.2

Quantity 2:

22234+48=?+18999-3494-1588+6622

?=1743

Comparing the quantities, x<y.

Q.7) A rectangular box is of length, height and breadth is to be made of tin. What is the area of tin sheet required if the box has lid also

Quantity 1:

Quantity 2:

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Explanation:

Quantity 1:

Total surface area=2(l×w+w×h+l×h)

=2(15×20+20×12+15×12)

=2(300+240+180)

=2(720)

=1440

Quantity 2:

Total surface area=2(l×w+w×h+l×h)

=2(17×19+19×9+9×17)

=2(323+171+153)

=2(647)

=1294

Hence Quantity 1 > Quantity 2

Q.8) Quantity 1:Find the area of the trapezium whose parallel sides are 15cm and 12 cm and the perpendicular distance between them is 19cm.

Quantity 2:Find the area of the quadrilateral piece of ground one of whose diagonal is 15 cm long and the perpendicular from the other two vertices are 12cm and 9cm respectively.

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Explanation:

Quantity 1:

Area of the Trapezium=1/2(Sum of the parallel sides)×(Perpendicular distance between them)

=1/2(12+15)×19

=1/2(27×19)

=256.5 cm2

Quantity 2:

d=15cm

h1=12cm

h2=9cm

Area=1/2×15(12+9)

=1/2×15(21)

=315/2=157.5 cm2

Hence Quantity 1>Quantity 2

Q.9) Quantity 1 : If side of a square is 5 cm. Then find its diagonal ?

Quantity 2: What is the diagonal if length & breadth of a rectangle is 4cm & 3cm ?

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Explanation:

Quantity I :

Diagonal of square = √2 × side = √2 ×5 = 7.07 cm

Quantity II :

Diagonal of rectangle = √(l2+b2 ) = √(42+ 32 ) = 5 cm

Hence Quantity 1> Quantity 2

Q.10) Quantity 1 : A person gets simple interest of Rs.1617 in 3yrs at 7% per annum. Find the principal ?

Quantity 2 : In 2 years a person gets Rs.19 more CI than SI at 5% as rate of interest. Find the principal?

a) Quantity 1 ≥ Quantity 2

b) Quantity 1 = Quantity 2 (or) No relation

c) Quantity 1 > Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 < Quantity 2

c)

Explanation:

Quantity I :

SI = (P×N×R)/100 = 1617

= (P×3×7)/100 = P = Rs. 7700

Quantity II :

CI – SI = P( R/100)2 = 19

= P(5/100)2 = P = Rs. 7600

Hence Quantity 1>Quantity 2