|
BACK
SOLVED PROBLEMS Complex Problems: 24.Six bells commence tolling together and toll at intervals of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many times do they toll together? Solution: To find the time that the bells will toll together we have to take L.C.M of 2,4,6,8,10,12 is 120. So,the bells will toll together after every 120 seconds i e, 2 minutes In 30 minutes they will toll together [30/2 +1]=16 times 25.The sum of two numbers is 15 and their geometric mean is 20% lower than their arithmetic mean. Find the numbers? a.11,4 b.12,3 c.13,2 d.10,5 Solution: Sum of the two numbers is a+b=15.If a=3 and a+b=15 then b=12. If a=12 and a+b=15 then b=3. Ans (b). 26.When we multiply a certain two digit number by the sum of its digits 405 is achieved. If we multiply the number written in reverse order of the same digits by the sum of the digits,we get 486.Find the number? a.81 b.45 c.36 d. none Solution: Let the number be x y. When we multiply the number by the sum of its digit 405 is achieved. (10x+y)(x+y)=405....................1 If we multiply the number written in reverse order by its sum of digits we get 486. (10y+x)(x+y)=486......................2 dividing 1 and 2 (10x+y)(x+y)/(10y+x)(x+y) = 405/486. 10x+y / 10y+x = 5/6. 60x+6y = 50y+5x 55x=44y 5x = 4y. From the above condition we conclude that the above condition is satisfied by the second option i e b. 45. Ans (b). 27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10? a.(x-2),(x-2)(x-3)(x-5) b.(x-2),(x-2)(x-3) c.(x-3),(x-2)(x-3)(x-5) d. none Solution: The given polynomials are x2-5x+6=0................1 x2-7x+10=0...............2 we have to find the factors of the polynomials x2-5x+6 and x2-7x+10 x2-2x-3x+6 x2-5x-2x+10 x(x-2)-3(x-2) x(x-5)-2(x-5) (x-3)(x-2) (x-2)(x-5) From the above factors of the polynomials we can easily find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5). Ans (c) 28.The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to? a.18 b.22 c.36 d. none Solution: Let the one digit numbers x,y,z Sum of all possible two digit numbers =(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z) Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22. 29.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed? Solution: Let the number be x.
3 - x
5 y - 1
8 z - 4
1 - 7
z=8*1+7=15 y=5z+4 = 5*15+4 = 79 x=3y+1 = 3*79+1=238
Now 8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2. 30.The arithmetic mean of two numbers is smaller by 24 than the larger of the two numbers and the GM of the same numbers exceeds by 12 the smaller of the numbers. Find the numbers? a.6,54 b.8,56 c.12,60 d.7,55 Solution: Let the numbers be a,b where a is smaller and b is larger number. The AM of two numbers is smaller by 24 than the larger of the two numbers. AM=b-24 AM of two numbers is a+b/2. a+b/2 = b-24 a+b = 2b-48 a = b-48...................1 The GM of the two numbers exceeds by 12 the smaller of the numbers GM = a+12 GM of two numbers is (ab)1/2Ans (a). 31.The sum of squares of the digits constituting a positive two digit number is 13,If we subtract 9 from that number we shall get a number written by the same digits in the reverse order. Find the number? a.12 b.32 c.42 d.52. Solution: Let the number be x y. the sum of the squares of the digits of the number is 13 x2+y2=13 If we subtract 9 from the number we get the number in reverse order x y-9=y x. 10x+y-9=10y+x. 9x-9y=9 x-y=1 (x-y)2 =x2+y2-2x y 1 =13-2x y 2x y = 12 x y = 6 =>y=6/x x-y=1 x-6/x=1 x2-6=x x2-x-6=0 x+2x-3x-6=0 x(x+2)-3(x+2)=0 x=3,-2. If x=3 and x-y=1 then y=2. If x=-2 and x-y=1 then y=-3. Therefore the number is 32. Ans (b). 32.If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number? a.18 b.39 c.49 d.28 Solution: Let the digit number be x y Given that if we add square of the digit in the tens place of a number to the product of the digits we get 52. x2+x y=52. x(x+y)=52....................1 Given that if we add the square of the digit in the unit's plac e to the product is 117. y2+x y= 117 y(x+y)=117.........................2 dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9 x/y=4/9 from the options we conclude that the two digit number is 49 because the condition is satisfied by the third option. Ans (c) 33.The denominators of an irreducible fraction is greater than the numerator by 2.If we reduce the numerator of the reciprocal fraction by 3 and subtract the given fraction from the resulting one,we get 1/15.Find the given fraction? Solution: Let the given fraction be x / (x+2) because given that denominator of the fraction is greater than the numerator by 2 1 – [(x – 1/(x+2))/3] = 1/15. 1 – (x2+2x-1) /3(x+2) = 1/15 (3x+6-x2-2x+1)/3(x+2) = 1/15 (7-x2+2x)/(x+2) = 1/5 -5x2+5x+35 = x+2 5x2-4x-33 = 0 5x2-15x+11x-33 = 0 5x(x-3)+11(x-3) = 0 (5x+11)(x-3) = 0 Therefore x=-11/5 or 3 Therefore the fraction is x/(x+2) = 3/5. 34.Three numbers are such that the second is as much lesser than the third as the first is lesser than the second. If the product of the two smaller numbers is 85 and the product of two larger numbers is 115. Find the middle number? Solution: Let the three numbers be x,y,z Given that z – y = y – x 2y = x+z.....................1 Given that the product of two smaller numbers is 85 x y = 85................2 Given that the product of two larger numbers is 115 y z = 115...............3 Dividing 2 and 3 x y /y z = 85/115 x / z = 17 / 23 From 1 2y = x+z 2y = 85/y + 115/y 2y2 = 200 y2 = 100 y = 10 35.If we divide a two digit number by the sum of its digits we get 4 as a quotient and 3 as a remainder. Now if we divide that two digit number by the product of its digits we get 3 as a quotient and 5 as a remainder . Find the two digit number? Solution: Let the two digit number is x y. Given that x y / (x+y) quotient=4 and remainder = 3 we can write the number as x y = 4(x+y) +3...........1 Given that x y /(x*y) quotient = 3 and remainder = 5 we can write the number as x y = 3 x*y +5...............2 By trail and error method For example take x=1,y=2 1............12=4(2+3)+3 =4*3+3 ! =15 let us take x=2 y=3 1..............23=4(2+3)+3 =20+3 =23 2.............23=3*2*3+5 =18+5 =23 the above two equations are satisfied by x=2 and y=3 Therefore the required number is 23. 36.First we increased the denominator of a positive fraction by 3 and then it by 5.The sum of the resulting fractions proves to be equal to 2/3. Find the denominator of the fraction if its numerator is 2. Solution: Let us assume the fraction is x/y First we increasing the denominator by 3 we get x/(y-3) Then decrease it by 5 we get the fraction as x/(y-5) Given that the sum of the resulting fraction is 2/3 x/(y+3) + x/(y-5) = 2/3 Given numerator equal to 2 2*[ 1/y+3 + 1/y-5] =2/3 (y-5+y+3) / (y-3)(y+5) =1/3 6y – 6 = y2-5y+3y-15 y2-8y-9 = 0 y2-9y+y-9 = 0 y(y-9)+1(y-9) = 0 Therefore y =-1 or 9. 37.If we divide a two digit number by a number consisting of the same digits written in the reverse order,we get 4 as quotient and 15 as a remainder. If we subtract 1 from the given number we get the sum of the squares of the digits constituting that number. Find the number? a.71 b.83 c.99 d. none Solution: Let the number be x y. If we divide 10x+y by a number in reverse order i e,10y+x we get 4 as quotient and 15 as remainder. We can write as 10x+y = 4(10y+x)+15......................1 If we subtract 1 from the given number we get square of the digits 10x+y = x2+y2.....................................2 By using above two equations and trail and error method we get the required number. From the options also we can solve the problem. In this no option is satisfied so answer is d. Ans (d) BACK |