Quant Test for IBPS PO Pre Part - 16

Ques 1.

12  34  98  286  ?

Solution :

12*3 – 2 = 34
34*3 – 4 = 98
98*3 – 8 = 286
286*3 – 16 = 842

Ques 2.

I. 2X2 + 11X + 12= 0  ;     II.2Y2 + 19Y + 45 = 0

Solution :

==> 2X^2+8X+3X+12 = 0

==> 2X(X+4)+3(X+4) = 0

==> (2X+3)(X+4) = 0

==> X = -3/2 and -4

==> 2Y^2+10Y+9Y+45 = 0

==>2Y(Y+5)+9(Y+5) = 0

==> (2Y+9)(Y+5) = 0

==> Y = -9/2 and -5

Ques 3.

Kiran and Ravi can do a piece of work in 5 and 7 days respectively. They start working alternatively starting from Ravi , then in how many days the work is completed

Solution :

1/7 + 1/5 = 12/35 (in 2 days)
==> completed in 4 days = 24/35
In the next day, 29/35 work get completed in 5 days, so remaining work will be completed by Kiran in 6/7 days

Ques 4.

√7921+√3969=? +52  

Solution :

==> 89+63 = ? +52

==> 152 = ?+52

Ques 5.

Siva borrows Rs.14000 at simple Interest from a lender. At the end of 3 years, he again borrows Rs.6000 and settled that amount after paying Rs.9230 as interest after 8 years from the time she made the first borrowing. what is the rate of interest?

Solution :

SI for Rs.14000 for 8 years= (14000*r*8)/100
==>Again borrowed=6000
==>SI = (6000*r*5)/100
Total interest= [(14000*r*8)/100] + [(6000*r*5)/100] = 9230
==>1120r+ 300r= 9230
==>1420r = 9230
==>r = 6.5%

Ques 6.

Krishna’s present age is 1.5 times of Gopal’s present age. If after 4 years, Krishna’s age will be twice of Gopal’s age 4 years ago. What is the difference between the present ages of Krishna and Gopal?

Solution :

Krishna = 1.5 * Gopal

Krishna + 4 = 2 * (Gopal – 4)
==>Krishna + 4 = 2Gopal– 8
==>Krishna + 12 = 2Gopal
==>1.5Gopal + 12 = 2Gopal
Gopal = 24
Krishna  = 24 * 1.5 = 36
Difference = 36 – 24 = 12

Ques 7.

Average age of women in a company is 32 years. If the average of the men in company is also added, the average becomes 31 years. If there were 40 men in the company with average age 30 years, how many women were there in the company ? 

Solution :

Let x Women in the Company 
==>32x + 40*30 = 31 (x+40)

==>X = 40*(31-30) = 40

Ques 8.

A rectangular courtyard 3.78 m long and 5.25 m broad is to be paved exactly with square tiles, all of the same size. What will be the minimum number of such tiles is?

Solution :

HCF of 378 and 525 =  21
Size of largest tile = 0.21* 0.21
Minimum Number of tiles = (3.78 * 5.25) / (0.21 * 0.21) = 450

Ques 9.

The price of a Slate and a pen is Rs.30. The slate was sold at a 20% profit and the pen at a loss of 5%. If the Shop keeper gains Rs.2 in the whole transaction, then how much is the cost price of slate ?

Solution :

20x/100 – 5(30-x)/100 = 2
==>20x + 5x = 200 + 150
==>25x = 350

Ques 10.

P,Q and R started a shop by investing Rs.54000, Rs.96000 and Rs.126000 respectively. At the end of the year the profits were distributed among them.If Q’s share of profit be Rs.38400, then the total profit was

Solution :

P:Q:R = 54:96:126 = 9:16:21
==>16x/46 = 38400
==>X = 38400*46/16 = 110400

Join 40,000+ readers and get free notes in your email

Let's block ads! (Why?)