# Numeric Entry Practice Test 3

#### Q.1

Find r when C(n,r) = 35 and P(n,r) = 840

• Explanation: From the given conditions, we get C(n,r) = n!/[r!(n-r)!] = 35...(1) P(n,r) = n!/(n-r)! = 840...(2) Dividing (2) by (1), we get n!/(n-r)!/n!/[r!(n-r)!] = 840/35 r! = 24 r! = 1*2*3*4 = 24 r = 4

#### Q.2

If (n+1)! = 30*(n-1)!, then find n.

• Explanation: (n+1)! = 30*(n-1)! (n+1)*n*(n-1)! = 30(n-1)! (n+1)*n = 30 n^2 + n - 30 = 0 n^2 +6n-5n - 30 =0 n(n+6) - 5(n+6) = 0 n = -6, 5 Since n cannot be negative, n = 5

#### Q.3

Find x if (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) are the vertices of an equilateral triangle in the first quadrant.

• Explanation: Let the vertices (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) be A, B and C respectively. Since the triangle is equilateral, we have AB = BC = CA AB^2 = BC^2 = CA^2 (3-0)^2+(sqrt(3)-0)^2 = (3-x)^2 + (2*sqrt(3) - sqrt(3))^2 = (x-0)^2 + (2*sqrt(3)-0)^2 9 + 3 = 9 - 2x + x^2 +3 = x^2 + 12 Hence, x^2+12 = 12 x = 0 [AB^2=AB*AB]

#### Q.4

C can complete the work in 18 days if he works alone. B takes 3 days lesser than C and A takes 5 days lesser than B to complete the work. On which day will they complete the work if C works for the initial two days only?

• Explanation: C takes 18 days to complete the work. B takes 3 days lesser. Hence, B takes 18-3 = 15 days. A takes 5 days lesser than B. Hence, A takes 15-5 = 10 days. Work done by A in one day = 1/10 Work done by B in one day = 1/15 Work done by C in one day = 1/18 Work done by the three of them in two days = 2(1/18+1/10+1/15) = 2(5+9+6)/90 = 2*20/90 = 4/9 Remaining work = 1-4/9 = 5/9 Work done by A and B in one day = 1/10+1/15 = (3+2)/30 =1/6 Time taken by A and B to complete the remaining work = 5/9*6/1 = 10/3 = 3.33 The work in completed on 2+3.33 = 6th day

#### Q.5

If A and B are two mutually exclusive events and P(A) = 2/3 and P(B) = 4/9, then find the probability of the occurrance of A and B together.

• Explanation: Since A and B are mutually exclusive events, they do not occur together at all. Hence, the required probability is zero.

#### Q.6

Find the diagonal of a cuboid whose volume is 144 cc and the dimensions of its base are 12cm and 4cm.

• Explanation: The length (l) and breadth (b) of the base are 12 cm and 4 cm respectively. Height (h) of the cuboid = volume/(l*b) = 144/(12*4) = 3 cm Diagonal of the cuboid = sqrt(l^2+b^2+h^2) = sqrt(12^2+4^2+3^2) = sqrt(144+16+9) = sqrt(169) = 13 cm The diagonal of the cuboid is 13 cm. [l^2=l*l]

#### Q.7

Find the value of k for which both the equations 12x^2+4kx+3=0 and (k+1)x^2-2(k-1)x+1=0 have equal real roots. [x^2=x*x]

• Explanation: Since the two equations have equal real roots, their discriminants are 0. For 12x^2+4kx+3=0, D = 0 D^2 = 0 b^2-4ac=0 (4k)^2 - 4*12*3 = 0 16k^2 - 144=0 k^2 = 144/16 = 9 k = -3, 3 For (k+1)x^2-2(k-1)x+1=0 b^2-4ac=0 [2*(k-1)]^2-4*(k+1)*1=0 4(k^2-2k+1)-4(k+1)=0 4[k^2-2k+1-k-1]=0 k^2 - 3k = 0 k(k-3)=0 k=0,3 Hence, we have k = 3 [k^2=k*k]

#### Q.8

What percent is the first number of the second if the first number is 120% of the third and the second number is 150% of the third?

• Explanation: Let the first, second and the third numbers be x, y and z respectively. x = 120% of z = 120/100*z y = 150% of z = 150/100*z x/y*100 = (120z/100)/(150z/100)*100 = 80% x is 80% of y.

#### Q.9

The difference between two numbers is 2. Each number is less than 14 and their sum is greater than 22. Find the greater number of the two.

• Explanation: Let the two numbers be x and y such that x < y. y ??? x = 2, x < 14, y < 14, x + y > 22 y = x + 2, x < 14, x + 2 < 14, x + x + 2 > 22 x < 14, x < 12, 2x + 2 > 22 x < 12, x +1 > 11 x < 12, x > 10 Hence, x can be 11 and the corresponding value of y is 13. The two numbers are 11 and 13.

#### Q.10

The sum of the squares of the two positive numbers is 68 and the square of their difference is 36. Find the sum of the numbers.

• Explanation: Let the numbers be x and y, x>y x^2+y^2=68 and (x-y)^2=36 (x-y)^2 = x^2+y^2-2xy 36 = 68-2xy 2xy=68-36=32 xy=32/2=16 (x+y)^2=x^2+y^2+2xy = 68+2*16 =68+32=100 x+y = sqrt(100) = 10 [x^2=x*x]

#### Q.11

The difference between the time when the lightening was seen and the time when the thunder was heard is 10 seconds. Sound covers a distance 330 meters in one second. Find the distance of the thundercloud from the point of observation in meters.

• Explanation: Distance = speed*time = 330*10 = 3300 meters

#### Q.12

Six points lie on a circle. How many cyclic quadrilaterals can be formed by joining the points?

• Explanation: Since, the points lie on a circle, all the possible quadrilaterals shall be cyclic. Number of quadrilateals = C(6,4) = 6!/(2!4!) = 6*5/2 = 15 Hence, 15 cyclic quadrilaterals can be drawn.

#### Q.13

When a number 'a' is increased by 17, it equals 60 times its reciprocal. How many values of 'a' are possible?

• Explanation: According to the conditions, we have a+17=60*1/a a^2+17a-60=0 a^2 +20a-3a-60=0 a(a+20)-3(a+20)=0 (a-3)(a+20)=0 a=3, -20 When a = 3, a+17 = 3+17 = 20 = 60*1/3 When a = -20, a+17 = -20+17 = -3 = 60*(-1/20) Hence, the two values of 'a' are valid. There are two possible values of 'a'. [a^2=a*a]

#### Q.14

The tax on a commodity decreases by 10% and the consumption increases by 20%. What is the percentage increase in the revenue?