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# Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

9(x+10)= x² X=-6 ,15 X=15 (as -6 not possible)

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

9(x+10)= x² X=-6 ,15 X=15 (as -6 not possible)

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

9(x+10)= x² X=-6 ,15 X=15 (as -6 not possible)

Tutor

Let's assume the marks obtained by Ravita are x. According to the statement, 9(x+10) = x^2 9x +90 = x^2 x^2-9x-90 = 0 which is in the form of ax^2+bx+c=0 x=(-b±?(b^2-4ac))/2a, x=9±?(81+360))/2, x=(9+21)/2, x= 15 if x = (9-21)/2, x= -6 Assuming the positive narks scored, she scored 15 marks.

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Formula of A P
an = a+(n-1)d Where an=last term a=first term d=common difference n=number of term Note: last term is denoted by an or l Sn=n/2[2a+(n-1)d Sn=sum of all terms

Manjeet Khajuria | 19/10/2022

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