# Numeric Entry Practice Test 4

#### Q.1

From a rectangular cardboard of length 21 cm and breadth 12 cm, a square whose diagonal is of length 3*sqrt(2) cm is cut out. Find the area of the remaining cardboard.

• Explanation: Area of the cardboard = length*breath = 21*12 = 252 sq.cm. Diagonal of a square of side x cm is equal to x*sqrt(2) cm. Hence, side of square = 3 cm Area of square = side*side = 3*3 = 9 sq.cm. Area of the remaining cardboard = 252-9 = 243 sq.cm.

#### Q.2

The height of a cone is trebled. The radius of the cone is doubled. By how many times will the volume of the cone increase?

• Explanation: Let the original height be h and radius of the base be r. Original volume will be = (1/3)*pi*r^2*h New height will be 3h New radius will be 2r New volume will be = (1/3)*pi*(2r)^2*(3h) = (1/3)*pi*r^2*h = (1/3)*pi*r^2*h*12 New volume = 12*original volume Hence, the volume increases 12 times. [pi=22/7, r^2=r*r]

#### Q.3

A rectangular reservoir is 120 m long and 75 m high. The cross-sectional area of the pipe through which water flows is 0.04 sq.m. and the water level in the reservoir rises 2.4 meters in 18 hours. Find the speed of water flowing into the reservoir.

• Explanation: Cross-sectional area of the pipe = 0.04 sq.m. Volume of water put into the reserviour in 18 hours = 120*75*2.4 = 21600 cubic m Volume of water put into the reserviour per hour = 21600/18 cubic m/hr Required speed = Volume/area = (21600/18)/0.04 = 30000 cubic m/hr

#### Q.4

P(x-1,3) : P(x,4) = 1: 9, find x.

• Explanation: P(x-1,3) : P(x,4) = 1: 9 (x-1)!/(x-1-3)! : x!/(x-4)! = 1:9 (x-1)!/(x-4)! : x(x-1)!/(x-4)! = 1:9 1:x = 1:9 x = 9

#### Q.5

If P(10,r) = 30240, then find r.

• Explanation: P(10,r) = 30240 10!/(10-r)! = 30240 10!/(10-r)! = 3024*10 10!/(10-r)! = 336*9*10 10!/(10-r)! = 42*8*9*10 10!/(10-r)! = 6*7*8*9*10 10!/(10-r)! = 10!/5! 10!/(10-r)! = 10!/(10-5)! Hence, r = 5

#### Q.6

Find the sum of the first 20 terms of the AP 1, 4, 7, 10...

• Explanation: The first term of the AP, a, is 1 and the common difference, d, is 3. The sum of n terms is given by Sn = (n/2)[2a+(n-1)d] Sum of 20 terms = (20/2)[2*1+(20-1)*3] = 10(2+19*3) = 10(59) = 590

#### Q.7

Tim buys two items at different rates for a total of Rs. 410 and sells one at a loss of 20% and the other at a gain of 25%. If both the items were sold at the same price, then find the cost price of the cheaper item.

• Explanation: Let CP and SP be cost price and selling price respectively. Let the CP of the first item be Rs.x and that of the second item be Rs.(410-x) Selling price of the first item SP = CP(100-loss%)/100 = x(100-20)/100 = 8x/10 Selling price of the second item SP = CP(100+gain%)/100 = (410-x)(100+25)/100 = (410-x)(125/100) Since both the items were sold at the same price, we have 8x/10 = (410-x)(125/100) 80x = 410*125 - 125x 80x + 125x = 51250 205x=51250 x = 250 The second item would cost 410-250 = 160 The cheaper item costs Rs.160

#### Q.8

Find the length of a parallel side of the trapezium if one of its parallel sides is 70 m long and its area is 2500 sq.m. The distance between the parallel sides is 40m.

• Explanation: Let the unknown length be x m. Area of trapezium = (1/2)*sum of parallel sides*distance between parallel sides 2500 = (1/2)*(70+x)*40 70+x = 2500*2/40 70+x = 125 x = 125-70 = 55 The other side of the trapezium is 55 m long.

#### Q.9

Find (b-a) when x^4+ax^2+bx+5 is exactly divisible be x^2+3x+2. [x^4=x*x*x*x]

• Explanation: x^2+3x+2=(x+1)(x+2) (x+1) and (x+2) should be the factors of x^4+ax^2+bx+5 Putting x = -1 and x = -2, we get x^4+ax^2+bx+5 = (-1)^4+a(-1)^2+b(-1)+5=0 1+a-b+5=0 a-b = -6 ...(1) x^4+ax^2+bx+5 = (-2)^4+a(-2)^2+b(-2) + 5=0 16 + 4a -2b+5=0 4a-2b = -21...(2) Multiplying (1) by 4 and subtracting (2) from it, we get 4a - 4b -4a +2b = -24 + 21 -2b = -3 b = 3/2 a = b-6 = 3/2-6 = (3-12)/2 = -9/2 b -a = 3/2+9/2 = 12/2 = 6

#### Q.10

x+1/x = 5. Find x^3+1/x^3. [x^3=x*x*x]

• Explanation: x+1/x = 5 (x+1/x)^3 = 5^3 x^3+1/x^3+3(x+1/x) = 125 x^3+1/x^3 = 125 -3*5 =125-15 = 110 x^3+1/x^3 = 110

#### Q.11

Pam lent Rs. 1800 to Ritu. She also lent Rs 2250 to Tammy for four years. They both returned the same interest charged at the same rate of interest of 3% per annum. For how many years did Ritu borrow the money?

• Explanation: Let P, R and T be the principle, rate and time for the simple interest SI SI = P*R*T/100 Since the SI is the same for both of them, we have 1800*3*T/100 = 2250*3*4/100 T = 2250*3*4/(1800*3) = 5 The money was lent for 5 years

#### Q.12

The length of the tangent drawn from an exterior point P to the circle is 63 cm. The point P is at a distance of 65 cm from the centre of the circle. Find the radius of the circle.

• Explanation: Let the centre of the circle be the point O and let the point at which the tangent drawn from point P meets the circle be T. PTO will be a right triangle right angled at T. PT = 63 cm and PO = 65 cm. Applying Pythagoras theorem to the triangle PTO and squaring both the sides, we get PO^2 = PT^2+TO^2 65^2=63^2+TO^2 4225 = 3969 + TO^2 TO^2 = 4225-3969 = 256 TO = 16 cm The radius of the circle is 16 cm. [PO^2=PO*PO]

#### Q.13

A hollow cylinder of height 3m is melted to form a solid cylinder of the same height. The external and internal radii of the hollow cylinder are 29 cm and 20 cm respectively. Find the radius of the solid cylinder.

• Explanation: Volume of cylinder = pi*r^2*h, where r is the radius and h is the height. Internal radius = 20 cm and external radius = 29 cm Volume of metal used = pi*(29)^2*h - pi(20)^2*h = pi*h[841-400] = pi*h*441 Volume of solid cylinder = pi*h*441 = pi*h*R^2, where R is the radius of the solid cylinder R^2 = 441 R = sqrt(441) = 21 The radius of the solid cylinder is 21 cm. [pi=22/7, r^2=r*r]

#### Q.14

A shopkeeper sells each book for Rs 1134 after giving a discount of 19% on the marked price. Had he sold the books at the printed price, he would have earned a profit of 40%. Find the cost price of each book.